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Differential Equations Unit 1: First-order equations

First-order differential equations involve only y and y′. Most you'll see are separable, linear or exact, and each type has its own method.

Covers: Separable, linear (integrating factor), exact equations, slope fields

Study guide

A first-order differential equation relates y, its first derivative and x. Slope fields show the equation visually: short segments with slope f(x, y) at many points, which solution curves follow. Equilibrium solutions are horizontal lines where the slope is zero.

Separable equations can be written dy/dx = g(x)h(y). Separate variables to get dy/h(y) = g(x) dx, integrate both sides, include one constant C, and use an initial condition to find C. Dividing by h(y) can lose constant solutions where h(y) = 0, so check for them separately.

Linear first-order equations have the standard form y′ + P(x)y = Q(x). Multiply through by the integrating factor μ(x) = e^(∫P dx). The left side becomes the derivative of μy, so (μy)′ = μQ; integrate both sides and divide by μ to get y. Always put the equation in standard form first so the coefficient of y′ is 1.

An exact equation M dx + N dy = 0 satisfies ∂M/∂y = ∂N/∂x. Then there's a potential function F with F_x = M and F_y = N, and the solution is F(x, y) = C. Find F by integrating M with respect to x, then differentiating with respect to y and matching N to find the missing function of y.

The existence and uniqueness theorem guarantees a unique solution through a point when f and ∂f/∂y are continuous near it.

An autonomous equation y′ = f(y) has equilibria where f(y) = 0. A phase line marks these on a number line with arrows showing where f is positive (y increases) or negative (y decreases). An equilibrium is stable if the arrows on both sides point toward it, unstable if both point away, and semistable if one points toward it and one away. An isocline is a curve along which every slope-field segment has the same slope: f(x, y) = m. The nullcline f(x, y) = 0 is the isocline of horizontal slopes, so a solution has its local maxima and minima where it crosses the nullcline.

For y′ = f(y), the chain rule gives y″ = f′(y)·f(y). Solutions are concave up where this product is positive and concave down where it's negative, and inflection points can occur where f′(y) = 0. For a linear equation y′ + P(x)y = Q(x), if P and Q are continuous on an open interval containing x₀, the solution with y(x₀) = y₀ exists and is unique on that whole interval. So the interval runs from x₀ out to the nearest point where P or Q is discontinuous.

Near an equilibrium y* of y′ = f(y), the equation behaves like the linear equation u′ = f′(y*)u, where u = y − y*. If f′(y*) 0 it is unstable.

Key ideas

Formulas and rules

Integrating factor: μ(x) = e^(∫P(x) dx)
Then (μy)′ = μQ
Exact test: ∂M/∂y = ∂N/∂x

Common mistakes

Words to know

Standard form of a first-order linear DE
y′ + P(x)y = Q(x)
Integrating factor
μ(x) = e^(∫P dx)
Test for an exact equation M dx + N dy = 0
∂M/∂y = ∂N/∂x
Separable equation form
dy/dx = g(x)h(y)
Lost equilibrium solutions
Dividing by h(y) can drop constant solutions where h(y) = 0.
Differential equation
An equation relating a function to its derivatives.
Order of a DE
The highest derivative that appears.
Slope field (direction field)
Segments showing the slope f(x, y) at many points.
Separable equation
dy/dx = g(x)h(y)
Exact equation
M dx + N dy = 0 with ∂M/∂y = ∂N/∂x.
Potential function (exact equations)
F with F_x = M and F_y = N; the solution is F = C.
Existence and uniqueness theorem
If f and ∂f/∂y are continuous near a point, a unique solution passes through it.

Sample practice questions

25 more questions like these, from level 1 to 5, are in the practice set.

Question 1 (level 1)

Order of y″ + 3y = x.

  1. 1
  2. 2
  3. 0
  4. 3
Show the answer and steps

Answer: B) 2

  1. The highest derivative is y″.
  2. Order 2.
Question 2 (level 2)

Solve y′ = 2y with y(0) = 3.

  1. y = 3e^(2x)
  2. y = e^(2x) + 3
  3. y = 2e^(3x)
  4. y = 3eˣ
Show the answer and steps

Answer: A) y = 3e^(2x)

  1. y′ = ky has solution y = Ce^(kx).
  2. k = 2, and y(0) = 3 gives C = 3.
  3. y = 3e^(2x).
Question 3 (level 3)

Integrating factor for y′ + 2y = x.

  1. e^(−2x)
  2. 2x
  3. e^(2x)
  4. eˣ
Show the answer and steps

Answer: C) e^(2x)

  1. Standard form y′ + P(x)y = Q(x) with P = 2.
  2. μ = e^(∫2 dx) = e^(2x).